At 80° textC , the vapour pressure of pure liquid 'A' is text520 textmm textHg and that of pure liquid 'B' is text1000 textmm textHg text. If a mixture solution of 'A' and 'B' boils at 80° textC and 1 atm pressure, the amount of 'A' in the mixture is: (. text1 textatm text=760 textmm textHg .)

Question

At 80° textC , the vapour pressure of pure liquid 'A' is text520 textmm textHg and that of pure liquid 'B' is text1000 textmm textHg text. If a mixture solution of 'A' and 'B' boils at 80° textC and 1 atm pressure, the amount of 'A' in the mixture is: (. text1 textatm text=760 textmm textHg .) (A) 52 mole percent (B) 34 mole percent (C) 48 mole percent (D) 50 mole percent

Tardigrade Admin · Accepted Answer

P T = P A ° X A + P B ° X B Mixture solution boil at 1 atm =760 mm Hg = total pressure of solution. 760=(520 X ) A +1000(1-( X ) A ) ⇒ 760=520 X A +1000-1000 X A ⇒ 480 X A =240 X A =0.5, mole percentage of A =50 %

Q. At $8 0^{^{\circ}} C$ , the vapour pressure of pure liquid 'A' is $520 mm Hg$ and that of pure liquid 'B' is $1000 mm Hg .$ If a mixture solution of 'A' and 'B' boils at $8 0^{^{\circ}} C$ and 1 atm pressure, the amount of 'A' in the mixture is:
$(1 atm =760 mm Hg)$

52 mole percent

34 mole percent

48 mole percent

50 mole percent

$P_{T =} P_{A}^{\circ} X_{A} + P_{B}^{\circ} X_{B}$
Mixture solution boil at $1 a t m = 760 mm H g =$ total pressure of solution. $760 = (520 X)_{A} + 1000 (1 - (X)_{A})$
$\Rightarrow 760 = 520 X_{A} + 1000 - 1000 X_{A} \Rightarrow 480 X_{A} = 240$
$X_{A} = 0.5$ ,
mole percentage of $A = 50%$

Q. At 80∘C , the vapour pressure of pure liquid 'A' is 520mmHg and that of pure liquid 'B' is 1000mmHg. If a mixture solution of 'A' and 'B' boils at 80∘C and 1 atm pressure, the amount of 'A' in the mixture is: (1atm=760mmHg)